When adjacent spot colours are printed through separate screens, how should the design handle a small expected alignment tolerance at a shared boundary?
A Remove one colour separation
B Leave a deliberate white gap
C Add controlled trapping overlap
D Double the screen circumference
Answer: C. Add controlled trapping overlap. A small intended overlap helps prevent a visible gap when normal registration error shifts one separation.
A print belt moves 1.8 m wide cloth at 24 m/min. Effective utilization is 75%. What average area is printed per hour?
A 2,592 m²/h
B 1,296 m²/h
C 432 m²/h
D 1,944 m²/h
Answer: D. 1,944 m²/h. Nominal throughput is 1.8 × 24 × 60 = 2,592 m²/h. Multiply by 0.75.
Full coverage at 600 × 600 dots/in² uses one 8 pL droplet at every dot. What theoretical ink volume lands per square inch?
A 0.288 µL
B 288 µL
C 28.8 µL
D 2.88 µL
Answer: D. 2.88 µL. There are 360,000 dots. Multiply by 8 pL to obtain 2,880,000 pL, or 2.88 µL.
Immediately after curing a dark motif looks acceptable, but a dry rub test removes pigment particles. What should be checked?
A Knitted wale density
B Binder film adhesion
C Repeat width in source file
D Darkroom exposure lamp power
Answer: B. Binder film adhesion. Dry rubbing can dislodge pigment if the binder film is undercured or adheres poorly to the cloth.
An uncompensated screen is 0.3% faster at its circumference than the cloth. Starting aligned, estimate slip after 20 m of fabric passes at that constant mismatch.
A 6 mm
B 0.6 mm
C 60 mm
D 600 mm
Answer: C. 60 mm. The accumulated relative displacement is 0.003 × 20 m = 0.060 m.
How much dye-free base lowers a 500 g paste from 6% dye to 4% dye by mass?
A 250 g
B 125 g
C 750 g
D 500 g
Answer: A. 250 g. There are 30 g of dye. At 4%, total paste mass must be 750 g. Add 250 g of base.
While cloth runs at 30 m/min, curing requires at least 50 seconds in the heated zone. What minimum path length is needed?
A 50 m
B 36 m
C 25 m
D 15 m
Answer: C. 25 m. Fifty seconds is 5/6 minute. Multiply by 30 m/min to obtain 25 m.
Rotary printing needs 45 minutes of setup and then runs at 30 m/min. Digital printing needs 5 minutes of setup and then runs at 10 m/min. At what order length do their total times match?
A 600 m
B 400 m
C 1,200 m
D 900 m
Answer: A. 600 m. Solve 45 + L/30 = 5 + L/10. This gives 40 = L/15 and L = 600 m.
To diagnose missing dots, solvent flush restores a group of nozzles temporarily and the defect returns after idle periods. Which mechanism is most consistent?
A Excess warp yarn count
B Ink drying at nozzles
C Fixed image mirror error
D Permanent belt width increase
Answer: B. Ink drying at nozzles. Idle time can allow drying or deposits at a nozzle. Temporary recovery after flushing supports a nozzle condition.
Wet paste add-on is 35 g/m², with 18% nonvolatile solids by mass. What dry add-on remains after the volatiles evaporate?
A 18.0 g/m²
B 28.7 g/m²
C 63.0 g/m²
D 6.3 g/m²
Answer: D. 6.3 g/m². Only 18% of the wet deposit remains, so dry add-on is 35 × 0.18.
At one shear setting, apparent viscosity is 0.12 Pa·s and shear stress is 24 Pa. What is the shear rate?
A 200 s⁻¹
B 288 s⁻¹
C 2.88 s⁻¹
D 20 s⁻¹
Answer: A. 200 s⁻¹. At a given condition, apparent viscosity equals shear stress divided by shear rate. Divide 24 by 0.12.
To support 600 m²/h of cloth carrying 22 g/m² of removable water, what evaporation rate must the dryer reach?
A 13.2 kg/h
B 27.3 kg/h
C 132 kg/h
D 1.32 kg/h
Answer: A. 13.2 kg/h. The hourly water input is 600 × 22 g, equal to 13.2 kg.
At 300 dpi, what nominal length is covered by three pixel pitches along the print direction?
A 0.085 mm
B 0.254 mm
C 0.762 mm
D 2.540 mm
Answer: B. 0.254 mm. Each pitch is 25.4/300 mm. Three pitches total 0.254 mm.
Independent printing stations each deposit 15 g/m² locally. Colour A covers 40%, B covers 30%, and their coverage overlaps by 10%. With additive deposition, what mean combined wet mass is applied per square metre?
A 10.5 g/m²
B 9.0 g/m²
C 7.5 g/m²
D 15.0 g/m²
Answer: A. 10.5 g/m². Each station deposits across its own area. Add 15 × 0.40 and 15 × 0.30, including both deposits where they overlap.
If one station prints 101 repeats over a distance where another prints 100 of equal nominal length, which fault is likely to accumulate?
A Constant visual colour match
B Even pigment curing
C Longitudinal phase drift
D Uniform fixed transverse shift
Answer: C. Longitudinal phase drift. Unequal repeat counts over the same length imply different effective pitches and accumulating registration error.
In a one-dimensional scan, a head travels 400 mm/s and ejects 8,000 droplets per second along a line. What is the nominal drop spacing?
A 5.00 mm
B 0.50 mm
C 0.05 mm
D 20.00 mm
Answer: C. 0.05 mm. Spacing equals head speed divided by drop frequency: 400/8,000 mm.
The back of a pile print appears much paler than the pile face even with correct alignment. Which comparison tests deeper penetration without changing pigment chemistry?
A Change only scale colour
B Vary the deposition amount
C Reverse only the artwork
D Change fibre blend and pressure
Answer: B. Vary the deposition amount. Changing the application amount while retaining the same chemistry helps test whether deposit and penetration cause the difference.
Finished motif width must be 150 mm. Shrinkage is 4% in that direction. How wide should the motif be before shrinkage?
A 144.00 mm
B 156.00 mm
C 150.04 mm
D 156.25 mm
Answer: D. 156.25 mm. The final size is 0.96 times the original print size. Divide 150 mm by 0.96.
Thickener stock is 10% solids and supplies all thickener solids in 8 kg of finished paste with 2.5% thickener solids. What stock mass is required?
A 3.2 kg
B 2.0 kg
C 0.80 kg
D 0.20 kg
Answer: B. 2.0 kg. The paste needs 0.20 kg thickener solids. At 10% solids, supply 2.0 kg of stock.
Suppose an ink head prints 1,024 pixels across one pass at 256 dpi. If each pass overlaps the previous pass by 64 pixels, what is effective advance between passes?
A 3.00 inches
B 0.25 inches
C 3.75 inches
D 4.00 inches
Answer: C. 3.75 inches. Each new pass adds 960 pixels of coverage. At 256 pixels/in, advance is 3.75 inches.
Colour A covers 48% of a design, colour B covers 32%, and 12% of the design is printed by both. What proportion remains entirely unprinted?
A 12%
B 20%
C 68%
D 32%
Answer: D. 32%. The covered fraction is 48% + 32% − 12% = 68%. The rest is unprinted.
Two lots have equal printed area and identical dry add-on. One uses 20% solids paste and the other 25% solids paste. What is the first lot’s wet deposit relative to the second’s?
A 1.05 times as high
B 1.25 times as high
C 0.80 times as high
D The same
Answer: B. 1.25 times as high. With equal dry add-on, wet deposit varies inversely with solids fraction. The ratio is 0.25/0.20.
Line speed rises from 20 to 25 m/min at constant heated path length. What happens to residence time?
A It falls by 20%
B It falls by 25%
C It remains unchanged
D It rises by 25%
Answer: A. It falls by 20%. Residence time is inversely proportional to speed. The new time is 20/25 of the old time.
Two pastes have the same apparent viscosity during the squeegee stroke but different recovery after the stroke. Which property helps predict resistance to later bleeding?
A Curing oven length only
B Screen roller circumference
C Dry cloth fibre count
D Low-shear viscosity recovery
Answer: D. Low-shear viscosity recovery. Once the stroke ends, recovery toward higher low-shear viscosity helps limit lateral spread.
The rotary screen rotates at 48 rpm and has an effective circumference of 0.64 m. With no slip, calculate cloth speed.
A 48.64 m/min
B 7.50 m/min
C 75.00 m/min
D 30.72 m/min
Answer: D. 30.72 m/min. One rotation advances 0.64 m. Multiply by 48 per minute.